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Natali5045456 [20]
2 years ago
8

If 8.0 g of NH4HS(s) is placed in a sealed vessel with a volume of 1.0 L and heated to 200 ∘C the reaction NH4HS(s) ⇌ NH3(g) + H

2S(g) will occur. When the system comes to equilibrium, some NH4HS(s) is still present. Which of the following changes will lead to a reduction in the amount of NH4HS(s) that is present, assuming in all cases that equilibrium is re-established following the change?
Adding more H2S(g) to the vessel
Adding more NH3(g) to the vessel
Decreasing the volume of the vessel
Adding more NH4HS(s) to the vessel
Increasing the volume of the vessel
Chemistry
1 answer:
PIT_PIT [208]2 years ago
7 0

Answer:

Increasing the volume of the vessel

Explanation:

By the Le Chatelier's principle, if a system in equilibrium suffer a variation that disturbs the equilibriu, the reaction shift in the way to minimize the pertubation and re-establish the equilibrium.

For a variation in pressure, when it increases, the reaction shift for the smallest of gas volume, and if decreases, the reaction will shift for the large gas volume. So, for the reaction given, the products have the large amount of gas, so by decreasing the pressure, more products will be formed, and the amount of NH₄HS will reduce. To decrease the pressure, we can increase the volume of the vessel: for the ideal gas equation (PV= nRT), pressure and volume are indirectly proportional.

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Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t
mars1129 [50]

Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0
2 years ago
Draw the best lewis structure for bro4- and determine the formal charge on bromine
Svetach [21]

Answer :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule. The electrons are represented by dot.

The given molecule is, perbromate ion.

Bromine has '7' valence electrons and oxygen has '6' valence electron.

Therefore, the total number of valence electrons in perbromate ion, BrO_4^- = 7 + 4(6) + 1 = 32

According to Lewis-dot structure, there are 14 number of bonding electrons and 18 number of non-bonding electrons.

Formula for formal charge :

\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}

\text{Formal charge on }Br=7-0-\frac{14}{2}=0

The Lewis-dot structure of perbromate ion is shown below.

7 0
2 years ago
Read 2 more answers
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):

I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).

See figure 1

I hope it helps!

7 0
2 years ago
The isotopes of an element all have the same __(atomic, mass) __number, but they have different __(atomic,mass)__numbers.
Vlad1618 [11]
<h2>The isotopes of an element all have the same __(atomic, mass) __number, but they have different __(atomic,mass)__numbers.</h2>

Explanation:

The isotopes of an element all have the same __atomic  number  __, but they have different __mass __numbers.

The isotopes have same atomic number that is :

  • Same number of electrons
  • Same number of protons
  • same electronic configuration
  • same valence electrons
  • same valency
  • same symbol

The isotopes have different mass number that is :

They differ in number of neutrons .

For example : Isotopes of hydrogen are : H₁¹ , H₁² , H₁³

                       isotopes of Oxygen is : O¹⁶ , O¹⁷, O¹⁸

4 0
2 years ago
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